Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int v[100001][18][12], head[100001], cnt, f[100001][18], m, n, q, a, b, c,
    ct[100001], dep[100001];
struct edge {
  int t, next;
} ed[200003];
void adde(int f, int t) {
  ed[++cnt].t = t;
  ed[cnt].next = head[f];
  head[f] = cnt;
  ed[++cnt].t = f;
  ed[cnt].next = head[t];
  head[t] = cnt;
}
void uni(int a[], int b[], int c[]) {
  int h1 = 1, h2 = 1;
  for (int i = 1; i <= 10 && (a[h1] || b[h2]); i++) {
    if (!a[h1])
      c[i] = b[h2], h2++;
    else if (!b[h2])
      c[i] = a[h1], h1++;
    else if (a[h1] < b[h2])
      c[i] = a[h1], h1++;
    else
      c[i] = b[h2], h2++;
  }
}
void uni2(int a[], int b[]) {
  int c[11] = {0};
  uni(a, b, c);
  for (int i = 1; i <= 10; i++) a[i] = c[i];
}
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  f[u][0] = fa;
  for (int i = 1; i <= 17; i++)
    f[u][i] = f[f[u][i - 1]][i - 1],
    uni(v[u][i - 1], v[f[u][i - 1]][i - 1], v[u][i]);
  for (int i = head[u]; i; i = ed[i].next)
    if (ed[i].t != fa) dfs(ed[i].t, u);
}
int query(int x, int y, int c1[]) {
  if (dep[x] < dep[y]) x ^= y ^= x ^= y;
  for (int i = 17; i >= 0; i--)
    if (dep[x] - (1 << i) >= dep[y]) uni2(c1, v[x][i]), x = f[x][i];
  if (x == y) {
    uni2(c1, v[x][0]);
    int tmp = 0;
    for (int i = 1; i <= 10; i++)
      if (c1[i]) tmp++;
    return tmp;
  }
  for (int i = 17; i >= 0; i--)
    if (f[x][i] != f[y][i])
      uni2(c1, v[x][i]), uni2(c1, v[y][i]), x = f[x][i], y = f[y][i];
  uni2(c1, v[x][0]), uni2(c1, v[y][1]);
  {
    int tmp = 0;
    for (int i = 1; i <= 10; i++)
      if (c1[i]) tmp++;
    return tmp;
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) scanf("%d%d", &a, &b), adde(a, b);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &a);
    if (ct[a] < 10) v[a][0][++ct[a]] = i;
  }
  dfs(1, 0);
  for (int i = 1; i <= q; i++) {
    scanf("%d%d%d", &a, &b, &c);
    int s1[11] = {0};
    int s = query(a, b, s1);
    int s2 = min(s, c);
    printf("%d ", s2);
    for (int i = 1; i <= s2; i++) printf("%d ", s1[i]);
    printf("\n");
  }
}